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kiruha [24]
3 years ago
7

A few people who have a particular disease take part in a medical trial that tests the effect of a medicine on the disease. Half

the people are given medicine and the other half are given sugar pills, which have no effect on the disease. The medicine has a 60% chance of curing someone, but people who do not get the medicine still have a 10% chance of getting well. The probability that a person gets well is ___%, and the probability that the person was cured because of the medicine is ___%.
Mathematics
1 answer:
Rom4ik [11]3 years ago
3 0

Answer:

1.The probability that a person gets well is 70%,

2.the probability that the person was cured because of the medicine is 60%

Step-by-step explanation:

chance of curing by medicine=60%

chance of curing without medicine=10%

total chance of curing=60%+10%=70%

chance of person being cured by medicine=60%

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Exercise 5.2. Suppose that X has moment generating function
soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Step-by-step explanation:

Given:

moment generating function  of X as:

MX(t) = \frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}

a)  Now

Mean, E(X) = M_{X}'(t=0)

Thus,

M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}

or

M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}

also,

E(X^{2})=M_{X}''(t=0)

Thus,

M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}

or

M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}

Therefore,

Mean, E(X) = M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}

or

Mean, E(X) = - 0.5

and

E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}

or

E(X^{2}) = 9.5

also,

Variance(X) = E(X²) - E(X)²

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= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Therefore,

at x = 0, P(x) = \frac{1}{2}

at x= - 4 ,P(x) = \frac{1}{3}

at x = 5, P(x) = \frac{1}{6}

Thus,

E(X) =\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})

or

E(X) = - 0.5

also,E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})

E(X^{2})  = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

4 0
4 years ago
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